A circle passing through the point $mathrm{P}(alpha,beta)$ in the first quadrant touches the two coordinate axes at the points $A$ and $B$. The point $P$is above the line $AB$. The point $Q$on the line segment $AB$is the foot of perpendicular from $P$ on $AB$. If $PQ$ is equal to $11$ units, then value of $alpha beta$ is_________.

A circle passing through the point in the first quadrant

A circle passing through the point $\mathrm{P}(\alpha . \beta)$ in the first quadrant touches the two coordinate axes at the points $\mathrm{A}$ and $\mathrm{B}$ . The point $\mathrm{P}$ is above the line $\mathrm{AB}$ . The point $\mathrm{Q}$ on the line segment $A B$ is the foot of perpendicular from $\mathrm{P}$ on $\mathrm{AB}$ . If $\mathrm{PQ}$ is equal to 11 units, then value of $\alpha \beta$ is_________.
Option: 1
121

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

Let equation of circle is $(x-a)^{2}+(y-a)^{2}=a^{2}$
which is passing through $P(\alpha, \beta)$
then $(\alpha-a) 2+(\beta-a)^{2}=a^{2}$
$\alpha^{2}+\beta^{2}-2 \alpha a-2 \beta \alpha+a^{2}=0$

Here equation of $A B$ is $x+y=a$

Let $\mathrm{Q}\left(\alpha^{\prime}, \beta^{\prime}\right)$ be foot of perpendicular of $\mathrm{P}$ on $\mathrm{AB}$

$\frac{\alpha^{\prime}-\alpha}{1}=\frac{\beta^{\prime}-\beta}{1}=\frac{-(\alpha+\beta-\mathrm{a})}{2}$
$PQ^{2}=\left(\alpha^{\prime}-\alpha\right)+\left(\beta^{\prime}-\beta\right)=\frac{1}{4}(\alpha+\beta-a)^{2}+\frac{1}{4}(\alpha+\beta-a)^{2}$
$121=\frac{1}{2}(\alpha+\beta-\mathrm{a})^{2}$
$242=\alpha^{2}+\beta^{2}-2 \alpha a-2 \beta a+a^{2}+2 \alpha \beta$
$242=2 \alpha \beta$
$\Rightarrow \alpha \beta=121$